us5244 demonstrate calculus skills. gradients of functions many real life situations can be modelled...
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US5244US5244Demonstrate Calculus Skills
Gradients of FunctionsGradients of FunctionsMany real life situations can be modelled by straight lines or curves (functions)
e.g. The cost of hiring a taxi can be modelled by a straight line where the slope (gradient) represents the cost per kilometre
e.g. For the distance travelled by a ball, the gradient represents the velocity of the ball
e.g. For a graph of a roller coaster’s profile, the gradient can represent its steepness at any particular point.
Distance (km)
Cost ($)
Time (s)
Height (m)
Height (m)
Time (s)
Gradients FunctionsGradients Functions
Below is the function y = x2
x-4 -2 2 4
y
2
4
6
8
10
To find the gradient at any particular point you need to calculate the gradient of the tangent to that point.
x Gradient
-3
-2
-1
0
1
2
3
The formula to find the gradient at any point is the gradient function.
The gradient function of y = x2
= 2x
-6
-4-2
0
2
4
6
Finding Gradients Functions (Differentiating)Finding Gradients Functions (Differentiating)
Through calculating gradients of other functions, the following results can also be found.
function gradient function
y = x3 dy/dx = 3x2
y = x4 dy/dx = 4x3
f(x) = x5 f’(x) = 5x4
f(x) = x6 f’(x) = 6x5
It is through these results that a pattern emerges:
If the function is written y = the gradient function is dy/dx = If the function is written f(x) = the gradient function is f’(x) =
If y = xn then dy/dx = nxn-1
If f(x) = xn then f’(x) = nxn-1
Two other important results can also be established
If f(x) = axn then f’(x) = n×axn-1
If f(x) = g(x) + h(x) then f’(x) = g’(x) + h’(x)
e.g. Find the gradient functions (differentiate) of the following y = x3 + 4x - 5 f(x) = 2x4 – 5x3 + 3x2 - 4
dy/dx = + 4 4×2x4-1
f’(x) = 8x3
3x2 f’(x) = – 3×5x3-1 + 2×3x2-1
– 15x2 + 6x
Sketching Gradients FunctionsSketching Gradients FunctionsThese sketches show how the gradient changes for a function1. Gradients of Straight Lines
x-4 -2 2 4
y
-4
-2
2
4
With a straight line, the gradient is always constant.
For the above example, the gradient is always 2 so we draw a horizontal line through 2.
x-4 -2 2 4
y
-4
-2
2
4
For the above example, the gradient is always -3 so we draw a horizontal line through -3.
2. Gradients of Quadratics (Parabolas)The gradient function of a quadratic is always a straight lineIf the coefficient of x2 is positive, the gradient function is positive.If the coefficient of x2 is negative, the gradient function is negative.
x-4 -2 2 4
y
-4
-2
2
4
- Look for when the gradient is 0 and mark the point on the x-axis
- The line goes above the x-axis where the quadratic has a positive slope, and below where it is negative
x-4 -2 2 4
y
-4
-2
2
4
- Mark the point on the x-axis where the gradient is 0
- The line goes above the x-axis where the quadratic has a positive slope, and below where it is negative
3. Gradients of CubicsThe gradient function of a cubic is always a quadratic (parabola)If the cubic goes from bottom to top, the gradient function is positiveIf the cubic goes from top to bottom, the gradient function is negative
x-4 -2 2 4
y
-4
-2
2
4
x-4 -2 2 4
y
-4
-2
2
4
- Look for when the gradient is 0 and mark the points on the x-axis
- Look for when the gradient is 0 and mark the points on the x-axis
- The parabola goes above the x-axis where the cubic has a positive slope, and below where it is negative
- The parabola goes above the x-axis where the cubic has a positive slope, and below where it is negative
Antidifferentiation or IntegrationAntidifferentiation or IntegrationThis is the reverse process to differentiation
e.g. 2x dx = x2
3x2 dx = x3
4x3 dx = x4
We know however, that when we differentiate, any number (constant) disappears, therefore when integrating we must always add in a constant (c)
In general: xn dx = xn + 1 + c n + 1
e.g. 7x6 dx =
(9x2 – 6x + 3) dx =
(2x3 + 3x2 - 8x - 5) dx =
7x7 7
= x7 + c
9x3
3 = 3x3 - 3x2 + 3x + c
2x4
4 = 1x4 + x3 - 4x2 + c
2
+ c
+ c
+ c
+ 3x- 6x2
2
+ c
+ c
+ 3x3
3
- 8x2 2
+ c